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2 years ago

Is the reflection in the x-axis, y-axis or neither

Mathematics

1 answer:

kykrilka [37]2 years ago

4 0

Reflection over the y-axis

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One month Abdul rented 3 movies and 5 video games for a total of . The next month he rented 6 movies and 2 video games for a tot

MrMuchimi

Answer:

Video Games = $5.25

Movies = $1.25

Step-by-step explanation:

(3x + 5y = 30)2

6x + 2y = 18

Video Games (y) = 5.25

6x + 10y = 60

(8y)/8 = (42)/8

6x + 2(5.25) = 18

6x + 10.5 = 18

6x = 7.5

x = 1.25

Double Check by plugging in the numbers

7 0

2 years ago

Plz..... Solve this questions for me...

faust18 [17]

$6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================$

$7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================$

$8.\\5^a\times5^b=5^{11}\\\\a)\\\text{use}\ a^n\times a^m=a^{n+m}\\\\5^a\times5^b=5^{a+b}\\\\5^{a+b}=5^{11}\Rightarrow a+b=11\\\\11\ \text{is odd. The sum of even and odd is odd. Therefore}\ a\ \text{and}\ b\ \text{can't be even.}\\--------------------\\\\b)\\5\ \text{solutions}\\\\11=5+6\\11=4+7\\11=3+8\\11=2+9\\11=1+10$

3 0

2 years ago

The mean one-way commute to work in Chowchilla is 7 minutes. The standard deviation is 2.4 minutes, and the population is normal

adell [148]

Answer:

The answer is below

Step-by-step explanation:

Given that:

The mean (μ) one-way commute to work in Chowchilla is 7 minutes. The standard deviation (σ) is 2.4 minutes.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}$

a) For x < 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

From normal distribution table, P(x < 2) = P(z < -2.08) = 0.0188 = 1.88%

b) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(2 < x < 11) = P(-2.08 < z < 1.67 ) = P(z < 1.67) - P(z < -2.08) = 0.9525 - 0.0188 = 0.9337

c) For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(x < 11) = P(z < 1.67) = 0.9525

d) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(2 < x < 5) = P(-2.08 < z < -0.83 ) = P(z < -0.83) - P(z < -2.08) = 0.2033- 0.0188 = 0.1845

e) For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(x < 5) = P(z < -0.83) = 0.2033

8 0

3 years ago

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0

2 years ago

7 cos x+1=6 secx solve for x

jonny [76]

7cos(x) + 1 = 6sec(x)
7cos(x) + 1 = 6/cos(x)
7cos^(x) + cos(x) = 6
7cos^(x) + cos(x) - 6 = 0
[7cos(x) - 6][cos(x) + 1] = 0
cos(x) = 6/7 , x = arccos(6/7) and
cos(x) = -1, x = 180

6 0

2 years ago